by sktsasus
Last Updated January 20, 2018 06:20 AM

**Solve for $h'(x)$ using the fundamental theorem of calculus.**

$$h(x) = \int _{-4}^{\sin\left(x\right)}\:\left(cos\left(t^4\right)+t\right)dt$$

I tried to do this by plugging in the $\sin(x)$ into both of the $t$'s and then tried to calculate the derivative of that.

This is the derivative I calculated:

$$-4\sin^3(x)\sin(\sin^4(x))\cos(x)+\cos(x)$$

But this is incorrect.

Any help?

$$h(x) = \int_{-4}^{\sin(x)} (\cos(t^4)+t) \, dt$$

Applying chain rule,

you should get $$\left(\cos (\color{blue}{\sin}^4\color{blue}{(x)})+\color{blue}{\sin(x)}\right) \frac{d}{dx}\sin (x)$$

Recall that in chain rule,

$$\frac{d}{dx}f(g(x))=f'(\color{blue}{g(x)})g'(x)$$

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